Question

Which of the following is not true for $N(Si{Me}_3{)}_3$?

• A. Three $s{p}^2$ orbitals are used for sigma bonding, giving a plane triangular structure
• B. The lone pair of electrons occupy a $p$-orbital at right angles to the plane triangle and this overlaps with the empty $p$-orbitals on each of the three $Si$ atoms resulting in $\pi$ bonding
• C. The $N-Si$ bond length is shorter than expected $N-Si$ bond length.
• D. It is a weaker Lewis base than tri methyl amine

Answer 1

The correct option is B The lone pair of electrons occupy a $p$-orbital at right angles to the plane triangle and this overlaps with the empty $p$-orbitals on each of the three $Si$ atoms resulting in $\pi$ bonding

The lone pair of nitrogen atom lie on the same plane as the empty $d$ orbitals of Si and so, $\pi$ - back bonding is formed. This reduces the availability of lone pair on N and hence it becomes a weaker Lewis base.

There is $p\pi -d\pi$ delocalization of lone pair of electrons on the nitrogen atom and the empty $d$-orbital of silicon. Because of this back bonding, $N(Si{Me}_3{)}_3$ is trigonal planar in shape and thus is $s{p}^2$ hybridised. There is a partial double bond character in each $N-Si$ bond and thus its bond length is shorter than expected.